Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main {
    static Scanner sc;

    private static void nextArr(int[] h) {
        for(int i = 0; i < h.length; i++) h[i] = sc.nextInt();
    }

    private static int nextInt() {
        return sc.nextInt();
    }

    private static void print(Object... vals) {
        for(int i = 0; i < vals.length; i++) {
            if (i != vals.length - 1) {
                System.out.print(vals[i] + ",");
            } else {
                System.out.print(vals[i]);
            }
        }
        System.out.println();
    }

    public static void main(String[] args) {
        sc = new Scanner(System.in);
        int n = sc.nextInt();
        double[] p = new double[n];
        for(int i = 0; i < n; i++) p[i] = sc.nextDouble();

        double[][] dp = new double[n + 1][n + 1];
        dp[1][1] = p[0];
        for(int i = 2; i <= n; i++) dp[1][i] = 0;
        for(int i = 2; i <= n; i++) {
            double prod = 1;
            for(int j = 1; j <= i; j++) prod *= (1 - p[j - 1]);
            for(int j = 1; j <= i; j++) {
                dp[i][1] += ((prod  * p[j - 1])/ (1 - p[j - 1]));
            }
        }
        for(int i = 2; i <= n; i++) {
            for(int j = 2; j<= n; j++) {
                dp[i][j] = p[i - 1] * dp[i - 1][j - 1] + (1 - p[i - 1]) * dp[i - 1][j];
            }
        }

        double ans = 0;
        for(int i = (n + 1) / 2; i <= n; i++) {
            ans += dp[n][i];
        }
        print(ans);
    }
}